Let $F=|\mathbf F|>0$ and choose an orientation of a <regular curve> following the <vector field>. Its <unit tangent vector> is $t=\sigma\mathbf F/F$, where $\sigma=\pm1$ is constant along a connected regular segment. <Arc length> differentiation along that <curve> is $d/ds=(\sigma\mathbf F/F)\cdot\nabla$. Therefore
$$
\frac{dt}{ds}
=\frac1F(\mathbf F\cdot\nabla)\left(\frac{\mathbf F}{F}\right)
=\frac{(\mathbf F\cdot\nabla)\mathbf F}{F^2}
-\frac{\mathbf F(\mathbf F\cdot\nabla)F}{F^3}.
$$
Taking the <cross product> with $t$ removes the term parallel to $\mathbf F$:
$$
t\times\frac{dt}{ds}
=\sigma\frac{\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F}{F^3}.
$$
Where the <Frenet frame> exists, $t\times t'=\kappa(t\times n)=\kappa b$. Thus
$$
\boxed{\frac{\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F}{F^3}
=\sigma\kappa b.}
$$
The printed sign corresponds to orientation along or against the field. Taking magnitudes gives the orientation-independent <curvature of an integral curve of a vector field>, $\kappa=|\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F|/F^3$. This <scalar> formula also applies at zero curvature, where $b$ itself may be undefined.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 3 / 11B / b / Solution

The <directional derivative> along the specified <vector field> gives
$$
(\mathbf F\cdot\nabla)\mathbf F=(x,y,z),\qquad
\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F=(2yz,-2xz,0).
$$
Use the magnitude formula for the <curvature of an integral curve of a vector field>:
$$
\boxed{\kappa(x,y,z)=\frac{2|z|\sqrt{x^2+y^2}}{(x^2+y^2+z^2)^{3/2}},\qquad (x,y,z)\ne(0,0,0).}
$$
It vanishes on the <plane> $z=0$ and on the $z$ axis away from the origin, consistent with straight field lines there. At the origin the <vector field> vanishes, so it fails the nowhere-zero hypothesis and this field-line curvature is not defined.

As an independent geometric check, an <integral curve of a vector field> satisfies $\dot x=x$, $\dot y=y$, $\dot z=-z$, giving $(x,y,z)=(ae^u,be^u,ce^{-u})$. Its velocity is $(x,y,-z)$ and acceleration $(x,y,z)$; the ordinary <curvature of a space curve> formula $|r'\times r''|/|r'|^3$ gives the same expression.

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 3 / 11B / c / Solution

{title2=$\kappa=|F\times(F\cdot\nabla)F|/|F|^3$}

For a <continuously differentiable> nonzero <vector field> $F$, the <curvature of a space curve> following its direction is
$$
\kappa=\frac{|F\times(F\cdot\nabla)F|}{|F|^3}.
$$
Take the <unit tangent vector> $t=\pm F/|F|$ and differentiate with respect to <arc length>. The derivative of $|F|^{-1}$ is parallel to $F$ and disappears from $t\times t'$. Since $t\cdot t'=0$, we have $|t\times t'|=|t'|=\kappa$, proving the formula. It remains valid with zero curvature, even where the usual <Frenet frame> is undefined.

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 Incoming links: Curvature of an integral curve of a vector field